php - Mysqli准备的语句选择不起作用，不返回错误

Question

我正在尝试将一些代码从常规mysqli选择查询切换到更安全的mysqli准备好的语句，但是无论我在过去几个小时中尝试了哪种方法，它都无法正常工作，在此代码中我有没有遗漏的东西吗？

function getProjectSites($selected){

    global $db;

    $result = $db->prepare("SELECT DISTINCT site_location, id FROM projects WHERE project_name_id = ? ORDER BY site_location");

    $result->bind_param('s', $selected);


    return $result->execute();    

}

$stmt = getProjectSites($selected);
$stmt->store_result();

/* Get the result */
$res = $stmt->get_result();

while ($row = $res->fetch_assoc()) {

//no rows are output

Answer 1

你可以试试这个->

function getProjectSites($selected){
    global $db;
    $result = $db->prepare("SELECT DISTINCT site_location, id FROM projects WHERE project_name_id = ? ORDER BY site_location");
    $result->bind_param('s', $selected);
    return $result->execute();    
}
$stmt = getProjectSites($selected);
$stmt->store_result(); 
// Fetch a record. Bind the result to a variable called 'value' and fetch. 
$stmt->bind_result($value) ; 
$res = $stmt->fetch() ; 
if($res) 
{ 
   //This is just an example you do your code here
   echo "data length is " . strlen($value);
}